How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 27$, where $x_1 \geq 0$, $x_2 \geq 0$, $x_3 \geq 0$, and $x_4 \geq 0$ are integers?
a) ${30 \choose 3}$
b) ${30 \choose 4}$
c) ${31 \choose 3}$
d) ${31 \choose 4}$
Solution: 2014 Fall Midterm - 8
Author: Michiel SmidQuestion
How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 27$, where $x_1 \geq 0$,
$x_2 \geq 0$, $x_3 \geq 0$, and $x_4 \geq 0$ are integers?
(a)
${30 \choose 4}$
(b)
${31 \choose 3}$
(c)
${31 \choose 4}$
(d)
${30 \choose 3}$
Solution
Suppose we have 27 blocks.
$x_1$ takes up 0 to 27 blocks. Same goes with $x_2, x_3$ and $x_4$
If we put dividers between each x, then we have 3 dividers. Since the dividers also take up space, then we can assume we have 27 spots + 3 spots for dividers.
We can put the dividers into any of these 30 spots and are looking for the number of combinations. We are choosing 3 spots from 30.
$\binom{30}{3}$